Liquid Solution

Step 1: Find the mole fraction of water.The solution has a mole fraction of solute $=0.02$.Mole fraction of water $=1$ minus mole fraction of solute.So, $\chi_{\text{water}} = 1-0.02 = 0.98$Step 2: Use Raoult's Law to relate vapour pressures.Raoult's law says: Vapour pressure of solution $=$ Vapour pressure of pure water $\times$ Mole fraction of water.$p_{\text{solution}} = p_{\text{water}} \times \chi_{\text{water}}$We know $p_{\text{solution}} = 34.65 \text{ mm Hg}$ and $\chi_{\text{water}} = 0.98$.So, $34.65 = p_{\text{water}} \times 0.98$Step 3: Solve for the vapour pressure of pure water.Divide both sides by 0.98 :$p_{\text{water}} = \frac{34.65}{0.98} = 35.36 \text{ mm Hg}$